# Newton polygons

This question is primarily to clear up some confusion I have about Newton polygons.

Consider the polynomial $x^4 + 5x^2 +25 \in \mathbb{Q}_{5}[x]$. I have to decide if this polynomial is irreducible over $\mathbb{Q}_{5}$.

So, I compute its Newton polygon. On doing this I find that the vertices of the polygon are $(0,2)$, $(2,1)$ and $(4,0)$. The segments joining $(0,2)$ and $(2,1)$, and $(2,1)$ and $(4,0)$ both have slope $-\frac{1}{2}$, and both segments have length $2$ when we take their projections onto the horizontal axis.

Am I correct in concluding that the polynomial $x^4 +5x^2 +25$ factors into two quadratic polynomials over $\mathbb{Q}_{5}$, and so is not irreducible?

I am deducing this on the basis of the following definition of a pure polynomial given in Gouvea's P-adic Numbers, An Introduction (and the fact that irreducible polynomials are pure):

A polynomial is pure if its Newton polygon has one slope.

What I interpret this definition to mean is that a polynomial $f(x) = a_nx^n + ... + a_0$ $\in \mathbb{Q}_{p}[x]$ (with $a_na_0 \neq 0$) is pure, iff the only vertices on its Newton polygon are $(0,v_p(a_0))$ and $(n, v_p(a_n))$. Am I right about this, or does the polynomial $x^4 + 5x^2+25$ also qualify as a pure polynomial?

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Do you really need to include $(2,1)$ on the list of vertices? IIRC you are supposed to exclude points on a line segment connecting two vertices. Otherwise you run into problems like the following. The polynomial $x^2+x+1\in\mathbb{Q}_5[x]$ is irreducible, because its roots are the third roots of unity, and they belong to an unramified quadratic extension of $\mathbb{Q}_5$. Yet according to your interpretation that polynomial would not be pure (vertices $(0,0)$, $(1,0)$ and $(2,0)$) and hence not irreducible? – Jyrki Lahtonen Apr 22 '12 at 19:46
@JyrkiLahtonen: I was not aware of the convention you mentioned, because when Gouvea defines a Newton polygon in his book, he does not mention that we should exclude points on a line segment joining two vertices. Also, that convention is not apparent from the definition of a Newton polygon given in Wikipedia. Is the convention you mention standard? Also, if you read my question again, you will see that I am not sure what the convention for a pure polynomial is. I gave the definition in Gouvea, and I wrote down what I interpret the definition to mean. I need to know if I am correct. – Rankeya Apr 22 '12 at 19:56
But, I see your point. It will be great if you could point me to a source that discusses Newton polygons with your conventions. – Rankeya Apr 22 '12 at 20:01
I don't know what the convention is. Sorry. I just wanted to point out that if you allow 3 collinear vertices, then that does not imply that the polynomial would be irreducible. In fact, I'm fairly sure that your example polynomial is irreducible. Find its zeros and check yourself, whether it has factors! – Jyrki Lahtonen Apr 22 '12 at 20:01
The zeros are sixth or third roots of unity scaled by $\sqrt5$, so it seems to me that including any single one of them necessarily creates a field extension with both $e>1$ and $f>1$. – Jyrki Lahtonen Apr 22 '12 at 20:20

There is no vertex at $(2,1)$. In my opinion, the right way to think of a Newton Polygon of a polynomial is as a closed convex body in ${\mathbb{R}}^2$ with vertical sides on both right and left. A point $P$ is only a vertex if there's a line through it touching the polygon at only one point. So this polynomial definitely is pure, and N-polygon theory does not help you at all. Easiest, I suppose, will be to write down what the roots are and see that any one of them generates an extension field of ${\mathbb{Q}}_5$ of degree $4$: voilà, your polynomial is irreducible.
The facts about the N-polygon are something you can prove on your own, and you'll see in the proof that you really do need a condition such as I specified. While I'm commenting, I might say that when there is a plot-point such as $(2,1)$ was here, lying on the boundary and coming from an actual monomial in the polynomial, it often happens that there’s a residue-field extension involved, as @Jyrki has already observed is the case here. – Lubin Apr 22 '12 at 21:16